It is being given that (232 + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number? A. (216 + 1) B. (216 - 1) C. (7 x 223) D. (296 + 1)

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232+1=233 is a prime number. So it has only two factors. One is 233 and 1. so all the options A, B, C, D are divisible by 1. A, B, C, D are not divisible by 233. A prime number has only two factors. One is 1, and another number is itself.
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Let 232 = x. Then, (232 + 1) = (x + 1). Let (x + 1) be completely divisible by the natural number N. Then, (296 + 1) = = (x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.
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Question seems ambiguous. Anyways. 233 is a prime number. The only numbers dividing it are 1 and 233. This seems trivial.
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All the options A. , B., C.,and D. are divisible by 1 which is a whole number.It can be calculated by finding the H.C.F. of all the given numbers and 233 which is equal to 1
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All have 1 as common factor.
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